To determine the molarity of a solution of sodium chloride after dilution, we can use the concept of conservation of moles during the dilution process. The number of moles of solute remains the same before and after dilution. We can use the dilution formula:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]
where:
- [tex]\( M_1 \)[/tex] = initial molarity (concentration) of the solution
- [tex]\( V_1 \)[/tex] = initial volume of the solution
- [tex]\( M_2 \)[/tex] = final molarity (concentration) of the solution
- [tex]\( V_2 \)[/tex] = final volume of the solution
Given:
1. The initial volume [tex]\( V_1 \)[/tex] is [tex]\( 750.0 \)[/tex] mL.
2. The initial molarity [tex]\( M_1 \)[/tex] is [tex]\( 6.0 \)[/tex] M.
3. The final volume [tex]\( V_2 \)[/tex] is [tex]\( 2.0 \)[/tex] L, which needs to be converted to milliliters (1 L = 1000 mL), so [tex]\( 2.0 \)[/tex] L = [tex]\( 2000.0 \)[/tex] mL.
Let's plug in these values into the dilution formula and solve for the final molarity [tex]\( M_2 \)[/tex]:
[tex]\[ (6.0 \, \text{M}) \times (750.0 \, \text{mL}) = M_2 \times (2000.0 \, \text{mL}) \][/tex]
Now, solve for [tex]\( M_2 \)[/tex]:
[tex]\[ M_2 = \frac{(6.0 \, \text{M}) \times (750.0 \, \text{mL})}{2000.0 \, \text{mL}} \][/tex]
[tex]\[ M_2 = \frac{4500.0 \, \text{M} \cdot \text{mL}}{2000.0 \, \text{mL}} \][/tex]
[tex]\[ M_2 = 2.25 \, \text{M} \][/tex]
Therefore, the molarity of the sodium chloride solution after diluting [tex]\( 750.0 \)[/tex] mL of a [tex]\( 6.0 \)[/tex] M solution to [tex]\( 2.0 \)[/tex] L is [tex]\( 2.25 \)[/tex] M.
